a is the answer because all of the other answers are wrtong
Answer:
ΔG° of reaction = -47.3 x
J/mol
Explanation:
As we can see, we have been a particular reaction and Energy values as well.
ΔG° of reaction = -30.5 kJ/mol
Temperature = 37°C.
And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:
The first step is to calculate the equilibrium constant for the reaction:
Equilibrium Constant K = ![\frac{[HPO4-2] x [ADP]}{ATP}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO4-2%5D%20x%20%5BADP%5D%7D%7BATP%7D)
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x
M
[ATP] = 1.2 x
M
[ADP] = 8.4 x
M
Let's plug in these values in the above equation for equilibrium constant:
K = ![\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2.1x10%5E%7B-3%7D%5D%20x%20%5B8.4x10%5E%7B-3%7D%5D%20%7D%7B%5B1.2%20x%2010%5E%7B-2%7D%5D%20%7D)
K = 1.47 x
M
Now, we have to calculate the ΔG° of reaction for the biological cell:
But first we have to convert the temperature in Kelvin scale.
Temp = 37°C
Temp = 37 + 273
Temp = 310 K
ΔG° of reaction = (-30.5
) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x
) + (-16810.68)
ΔG° of reaction = -47.3 x
J/mol
Answer:
561 g P₂O₃
Explanation:
To find the mass of P₂O₃, you need to (1) convert moles H₃PO₃ to moles P₂O₃ (via mole-to-mole ratio from equation coefficients) and then (2) convert moles P₂O₃ to grams P₂O₃ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the amount of sig figs in the given value.
Atomic Mass (P): 30.974 g/mol
Atomic Mass (O): 15.998 g/mol
Molar Mass (P₂O₃): 2(30.974 g/mol) + 3(15.998 g/mol)
Molar Mass (P₂O₃): 109.942 g/mol
1 P₂O₃ + 3 H₂O -----> 2 H₃PO₃
10.2 moles H₃PO₃ 1 mole P₂O₃ 109.942 g
---------------------------- x -------------------------- x ------------------- = 561 g P₂O₃
2 moles H₃PO₃ 1 mole
Explanation:
The given chemical equation is:

The rate of the reaction is 0.0352 M/s.
During the course of the reaction, the rate of reactants decreases, and the rate of products increases.
The rate of disappearance of B is shown below:
![rate=-\frac{1}{4} \frac{d[B]}{dt}](https://tex.z-dn.net/?f=rate%3D-%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
So, rate of change of B is :

Option C.